Let A be an $n \times k$ matrix with $W$ as the column space. Show that the solution space of $A(A^T) x = 0$ is given by $W^⊥$.

Question

How can I solve this question? $A^T$ refers to $A$ transposed and $W^⊥$ refers to the set of vectors orthogonal to W.

Answer 1

Denote by V the rowspace of $A$, i.e. the column space of $A^T$. Note that for $Av = 0$ we need that $v \in V^{\perp}$.

If $A^T x \neq 0$ then we have that $A^T x \in V$, which means $A^t x \not \in V^{\perp}$. Therefore we need $A^t x = 0$, which means that $x \in W^{\perp}$. We are done.

To see why for a matrix $A$ we have that $Ker(A) = V^{\perp}$ notice that the basis of $V$ is the set $(A^T_i)_{1 \leq i \leq n}$ (the columns of $A^T$). For $Ax = 0$ we need that $A^T_i \cdot x = 0$ for all $1 \leq i \leq n$, which implies that it is in $V^{\perp}$.

Answer 2

Let the solution space be $V=\{x\in R^n:AA^T x=0\}$. Let $A=(a_1, \cdots, a_k)$ and $W=Col (A)$. We want to show that $V=W^\perp$.

First, for any $w\in W^\perp$, by the definition, $a_i\cdot w=0$. Hence, $AA^Tw=0$, which implies that $W^\perp\subset V$.

Conversely, for any $x\in V$, we have $AA^Tx=0$, which implies $x^TAA^Tx=(A^Tx)\cdot (A^Tx)=0$, which in turn implies $a_i\cdot x=0$ for each $i$. Thus, $x$ is orthogonal to $span\{a_1, \cdots, a_k\}$. Hence,$x\in W^\perp$. We have $V\subset W^\perp$.